**Text from Educational Image:****Problem 7: Empirical and Molecular Formula Calculation**An unknown solvent (DHB) contains 40.02% carbon (C), 2.74% hydrogen (H), and 48.24% chlorine (Cl). If the molar mass of the compound is 293.87 g/mol, what are the empirical formula and molecular formula of this compound?**Explanation:**To find the empirical formula of a compound, we follow these steps:1. Convert the percentage composition to grams. Assume 100 grams of the compound, so the composition in grams will be the same as the percentage composition. - Carbon: 40.02 grams - Hydrogen: 2.74 grams - Chlorine: 48.24 grams2. Convert the grams to moles by dividing by the atomic mass of each element: - Moles of Carbon (C): 40.02 g / 12.01 g/mol = 3.33 moles - Moles of Hydrogen (H): 2.74 g / 1.01 g/mol = 2.71 moles - Moles of Chlorine (Cl): 48.24 g / 35.45 g/mol = 1.36 moles3. Find the simplest whole number ratio by dividing the moles of each element by the smallest number of moles calculated: - Carbon: 3.33 / 1.36 = 2.45 ≈ 2 - Hydrogen: 2.71 / 1.36 = 2.00 ≈ 2 - Chlorine: 1.36 / 1.36 = 1.00 ≈ 1 Therefore, the empirical formula is \( C_2H_2Cl \).4. Determine the molecular formula using the molar mass given (293.87 g/mol). First, calculate the molar mass of the empirical formula: - Molar mass of \( C_2H_2Cl \): - Carbon (C): 2 × 12.01 = 24.02 g/mol - Hydrogen (H): 2 × 1.01 = 2.02 g/mol - Chlorine (Cl): 1 × 35.45 = 35.45 g/mol - Total mol

Given that: mass % of C= 49.02 % mass % of Cl= 2.74 % mass % of H= 48.24 % To find: The empirical…

Answered: 7. An unknown solvent DHB, contains…

7. An unknown solvent DHB, contains 49.02% C, 2.74% H, and 48.24 % chlorine. If the molar mass is 293.87 g/mol, what is the empirical formula and molecular formula of this compound? vol t

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

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Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

**Text from Educational Image:**

**Problem 7: Empirical and Molecular Formula Calculation**

An unknown solvent (DHB) contains 40.02% carbon (C), 2.74% hydrogen (H), and 48.24% chlorine (Cl). If the molar mass of the compound is 293.87 g/mol, what are the empirical formula and molecular formula of this compound?

**Explanation:**

To find the empirical formula of a compound, we follow these steps:

1. Convert the percentage composition to grams. Assume 100 grams of the compound, so the composition in grams will be the same as the percentage composition.

- Carbon: 40.02 grams
- Hydrogen: 2.74 grams
- Chlorine: 48.24 grams

2. Convert the grams to moles by dividing by the atomic mass of each element:

- Moles of Carbon (C): 40.02 g / 12.01 g/mol = 3.33 moles
- Moles of Hydrogen (H): 2.74 g / 1.01 g/mol = 2.71 moles
- Moles of Chlorine (Cl): 48.24 g / 35.45 g/mol = 1.36 moles

3. Find the simplest whole number ratio by dividing the moles of each element by the smallest number of moles calculated:

- Carbon: 3.33 / 1.36 = 2.45 ≈ 2
- Hydrogen: 2.71 / 1.36 = 2.00 ≈ 2
- Chlorine: 1.36 / 1.36 = 1.00 ≈ 1

Therefore, the empirical formula is \( C_2H_2Cl \).

4. Determine the molecular formula using the molar mass given (293.87 g/mol). First, calculate the molar mass of the empirical formula:

- Molar mass of \( C_2H_2Cl \):
- Carbon (C): 2 × 12.01 = 24.02 g/mol
- Hydrogen (H): 2 × 1.01 = 2.02 g/mol
- Chlorine (Cl): 1 × 35.45 = 35.45 g/mol
- Total mol

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